Integration by Parts: Solving y = arcsinh(x)

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SUMMARY

The discussion focuses on using integration by parts to solve the integral of y = arcsinh(x). The key equation utilized is ∫(v du) = uv - ∫(u dv). Participants emphasize the importance of identifying 'u' and 'dv', with suggestions to use substitutions like x = sinh(a) and dx = cosh(a) da to simplify the integration process. A general tip shared is to familiarize oneself with the differentiation and integration of hyperbolic and trigonometric functions to enhance problem-solving efficiency.

PREREQUISITES
  • Understanding of integration by parts
  • Familiarity with hyperbolic functions
  • Knowledge of inverse hyperbolic functions
  • Basic substitution techniques in calculus
NEXT STEPS
  • Practice integration by parts with various functions
  • Learn hyperbolic function identities and their derivatives
  • Explore inverse hyperbolic function properties
  • Study substitution methods in calculus for complex integrals
USEFUL FOR

Students and educators in calculus, particularly those focusing on integration techniques, as well as anyone looking to strengthen their understanding of hyperbolic and inverse hyperbolic functions.

hex.halo
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Homework Statement



Use integration by parts to find:

y=... if dy=arcsinh(x) dx

Homework Equations



int(v.du)=uv-int(u.dv)

The Attempt at a Solution



I understand how to perform integration by parts. My problem here is, what are my 'v' and 'du'?
 
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\int sinh^{-1}x dx


well dv\neq sinh^{-1}x dx since to find v you'd need to integrate that and well obviously you can't do that. So u=sinh^{-1} and dv=1dx.
 
A good general tip is to take very hyperbolic function and every trig function and learn how to differentiate it or integrate it. That way you don't have to remember the tables of functions. If you like memorising tables though do it that way, but do both. :smile: Just a really good but obvious hint I picked up recently that I thought might be useful.

It's generally a good idea to try and split arc functions into there \frac{1}{\text{trig function}}[/tex] equivalents first, I don't know if it's just me but that seems to work out better more often than not? Clearly here the substitution becomes much easier when you do this, but I find it's a good general rule..?
 
Last edited:
hex.halo said:
I understand how to perform integration by parts. My problem here is, what are my 'v' and 'du'?

Hi hex.halo ! :smile:

You have to do a substitution first, and then integrate by parts!

Put x = sinha, dx = cosha da …

and you get … ? :smile:
 

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